forall x York Edition Solutions Booklet P.D. Magnus University at Albany, State University of New York Tim Button University of Cambridge Robert Trueman University of York This booklet contains model answers to the practice exercises found in forall x :Cambridge 2014-15 For several of the questions, there are mul- tiple correct possible answers; in each case, this booklet contains at most one answer. Answers are given in blue; please contact Rob Trueman at rob.trueman@york.ac.uk if you have accessibility requirements. © 2005–2021 by P.D. Magnus, Tim Button and Robert Trueman. Some rights reserved. This solutions booklet is based upon P.D. Magnus’s forall x (version 1.29), available at fecundity.com/logic, which was released under a Creative Com- mons license (Attribution-ShareAlike 3.0). You are free to copy this book, to distribute it, to display it, and to make derivative works, under the following conditions: (a) Attribution. You must give the original author credit. (b) Share Alike. If you alter, transform, or build upon this work, you may distribute the resulting work only under a license identical to this one. — For any reuse or distribution, you must make clear to others the license terms of this work. Any of these conditions can be waived if you get permission from the copyright holder. Your fair use and other rights are in no way affected by the above. — This is a human-readable summary of the full license, which is available on-line at http://creativecommons.org/licenses/by-sa/3.0/ In accordance with this license, Tim Button made changes to P.D. Magnus’s original text, and added new material, and he offers this solutions booklet for forall x :Cambridge 2014-15 under the same Creative Commons license. The most recent version is available at http://www.nottub.com/forallx.shtml. Also in accordance with this license., Robert Trueman has further modified Tim Button’s text. The resulting solutions booklet is intended to be used with forall x :York , and it is offered under the same Creative Commons license. The L A TEX source code is available on request from: rob.trueman@york.ac.uk Typesetting was carried out entirely in L A TEX2 ε The style for typesetting proofs is based on fitch.sty (v0.4) by Peter Selinger, University of Ottawa. Contents 2 Valid arguments 1 3 Other logical notions 2 6 Connectives 3 7 Sentences of TFL 8 11 Using truth tables 9 12 Semantic concepts 13 14 Basic rules for TFL 18 15 Additional rules for TFL 25 17 Derived rules 28 18 Proof-theoretic concepts 30 19 FOL: the basics 38 20 Identity 43 21 Definite descriptions 44 22 Sentences of FOL 48 24 Truth in FOL 49 26 Using interpretations 52 28 Basic rules for FOL 61 29 Conversion of quantifiers 74 30 Derived rules 80 31 Rules for identity 81 iii Valid arguments 2 Could there be: 1. A valid argument that has one false premise and one true premise? Yes. Example: the first argument, above. 2. A valid argument that has only false premises? Yes. Example: Socrates is a frog, all frogs are excellent pianists, therefore Socrates is an excellent pianist. 3. A valid argument with only false premises and a false conclusion? Yes. The same example will suffice. 4. A sound argument with a false conclusion? No. By definition, a sound argument has true premises. And a valid argument is one where it is impossible for the premises to be true and the conclusion false. So the conclusion of a sound argument is certainly true. 5. An invalid argument that can be made valid by the addition of a new premise? Yes. Plenty of examples, but let me offer a more general observation. We can always make an invalid argument valid, by adding a contradiction into the premises. For an argument is valid if and only if it is impossible for all the premises to be true and the conclusion false. If the premises are contradictory, then it is impossible for them all to be true (and the conclusion false). 6. A valid argument that can be made invalid by the addition of a new premise? No. An argument is valid if and only if it is impossible for all the premises to be true and the conclusion false. Adding another premise will only make it harder for the premises all to be true together. In each case: if so, give an example; if not, explain why not. 1 Other logical notions 3 A. Look back at the sentences G1–G4 in this section (about giraffes, gorillas and martians in the wild animal park), and consider each of the following: 1. G2, G3, and G4 Jointly consistent 2. G1, G3, and G4 Jointly inconsistent 3. G1, G2, and G4 Jointly consistent 4. G1, G2, and G3 Jointly consistent Which are jointly consistent? Which are jointly inconsistent? B. Could there be: 1. A valid argument, the conclusion of which is necessarily false? Yes: ‘1 + 1 = 3. So 1 + 2 = 4.’ 2. An invalid argument, the conclusion of which is necessarily true? No. If the conclusion is necessarily true, then there is no way to make it false, and hence no way to make it false whilst making all the premises true. 3. Jointly consistent sentences, one of which is necessarily false? No. If a sentence is necessarily false, there is no way to make it true, let alone it along with all the other sentences. 4. Jointly inconsistent sentences, one of which is necessarily true? Yes. ‘1 + 1 = 4’ and ‘1 + 1 = 2’. In each case: if so, give an example; if not, explain why not. 2 Connectives 6 A. Using the symbolisation key given, symbolise each English sentence in TFL. M : Those creatures are men in suits. C : Those creatures are chimpanzees. G : Those creatures are gorillas. 1. Those creatures are not men in suits. ¬ M 2. Those creatures are men in suits, or they are not. ( M ∨ ¬ M ) 3. Those creatures are either gorillas or chimpanzees. ( G ∨ C ) 4. Those creatures are neither gorillas nor chimpanzees. ¬ ( C ∨ G ) 5. If those creatures are chimpanzees, then they are neither gorillas nor men in suits. ( C → ¬ ( G ∨ M )) 6. Unless those creatures are men in suits, they are either chimpanzees or they are gorillas. ( M ∨ ( C ∨ G )) B. Using the symbolisation key given, symbolise each English sentence in TFL. A : Mister Ace was murdered. B : The butler did it. C : The cook did it. D : The Duchess is lying. E : Mister Edge was murdered. F : The murder weapon was a frying pan. 3 6. Connectives 4 1. Either Mister Ace or Mister Edge was murdered. ( A ∨ E ) 2. If Mister Ace was murdered, then the cook did it. ( A → C ) 3. If Mister Edge was murdered, then the cook did not do it. ( E → ¬ C ) 4. Either the butler did it, or the Duchess is lying. ( B ∨ D ) 5. The cook did it only if the Duchess is lying. ( C → D ) 6. If the murder weapon was a frying pan, then the culprit must have been the cook. ( F → C ) 7. If the murder weapon was not a frying pan, then the culprit was either the cook or the butler. ( ¬ F → ( C ∨ B )) 8. Mister Ace was murdered if and only if Mister Edge was not murdered. ( A ↔ ¬ E ) 9. The Duchess is lying, unless it was Mister Edge who was murdered. ( D ∨ E ) 10. If Mister Ace was murdered, he was done in with a frying pan. ( A → F ) 11. Since the cook did it, the butler did not. ( C ∧ ¬ B ) 12. Of course the Duchess is lying! D C. Using the symbolisation key given, symbolise each English sentence in TFL. E 1 : Ava is an electrician. E 2 : Harrison is an electrician. F 1 : Ava is a firefighter. F 2 : Harrison is a firefighter. S 1 : Ava is satisfied with her career. S 2 : Harrison is satisfied with his career. 6. Connectives 5 1. Ava and Harrison are both electricians. ( E 1 ∧ E 2 ) 2. If Ava is a firefighter, then she is satisfied with her career. ( F 1 → S 1 ) 3. Ava is a firefighter, unless she is an electrician. ( F 1 ∨ E 1 ) 4. Harrison is an unsatisfied electrician. ( E 2 ∧ ¬ S 2 ) 5. Neither Ava nor Harrison is an electrician. ¬ ( E 1 ∨ E 2 ) 6. Both Ava and Harrison are electricians, but neither of them find it sat- isfying. (( E 1 ∧ E 2 ) ∧ ¬ ( S 1 ∨ S 2 )) 7. Harrison is satisfied only if he is a firefighter. ( S 2 → F 2 ) 8. If Ava is not an electrician, then neither is Harrison, but if she is, then he is too. (( ¬ E 1 → ¬ E 2 ) ∧ ( E 1 → E 2 )) 9. Ava is satisfied with her career if and only if Harrison is not satisfied with his. ( S 1 ↔ ¬ S 2 ) 10. If Harrison is both an electrician and a firefighter, then he must be sat- isfied with his work. (( E 2 ∧ F 2 ) → S 2 ) 11. It cannot be that Harrison is both an electrician and a firefighter. ¬ ( E 2 ∧ F 2 ) 12. Harrison and Ava are both firefighters if and only if neither of them is an electrician. (( F 2 ∧ F 1 ) ↔ ¬ ( E 2 ∨ E 1 )) D. Give a symbolisation key and symbolise the following English sentences in TFL. A : Alice is a spy. B : Bob is a spy. C : The code has been broken. G : The German embassy will be in an uproar. 6. Connectives 6 1. Alice and Bob are both spies. ( A ∧ B ) 2. If either Alice or Bob is a spy, then the code has been broken. (( A ∨ B ) → C ) 3. If neither Alice nor Bob is a spy, then the code remains unbroken. ( ¬ ( A ∨ B ) → ¬ C ) 4. The German embassy will be in an uproar, unless someone has broken the code. ( G ∨ C ) 5. Either the code has been broken or it has not, but the German embassy will be in an uproar regardless. (( C ∨ ¬ C ) ∧ G ) 6. Either Alice or Bob is a spy, but not both. (( A ∨ B ) ∧ ¬ ( A ∧ B )) E. Give a symbolisation key and symbolise the following English sentences in TFL. F : There is food to be found in the pridelands. R : Rafiki will talk about squashed bananas. A : Simba is alive. K : Scar will remain as king. 1. If there is food to be found in the pridelands, then Rafiki will talk about squashed bananas. ( F → R ) 2. Rafiki will talk about squashed bananas unless Simba is alive. ( R ∨ A ) 3. Rafiki will either talk about squashed bananas or he won’t, but there is food to be found in the pridelands regardless. (( R ∨ ¬ R ) ∧ F ) 4. Scar will remain as king if and only if there is food to be found in the pridelands. ( K ↔ F ) 5. If Simba is alive, then Scar will not remain as king. ( A → ¬ K ) 6. Connectives 7 F. For each argument, write a symbolisation key and symbolise all of the sentences of the argument in TFL. 1. If Dorothy plays the piano in the morning, then Roger wakes up cranky. Dorothy plays piano in the morning unless she is distracted. So if Roger does not wake up cranky, then Dorothy must be distracted. P : Dorothy plays the Piano in the morning. C : Roger wakes up cranky. D : Dorothy is distracted. ( P → C ), ( P ∨ D ), ( ¬ C → D ) 2. It will either rain or snow on Tuesday. If it rains, Neville will be sad. If it snows, Neville will be cold. Therefore, Neville will either be sad or cold on Tuesday. T 1 : It rains on Tuesday T 2 : It snows on Tuesday S : Neville is sad on Tuesday C : Neville is cold on Tuesday ( T 1 ∨ T 2 ), ( T 1 → S ), ( T 2 → C ), ( S ∨ C ) 3. If Zoog remembered to do his chores, then things are clean but not neat. If he forgot, then things are neat but not clean. Therefore, things are either neat or clean; but not both. Z : Zoog remembered to do his chores C : Things are clean N : Things are neat ( Z → ( C ∧ ¬ N )), ( ¬ Z → ( N ∧ ¬ C )), (( N ∨ C ) ∧ ¬ ( N ∧ C )). G. We symbolised an exclusive or using ‘ ∨ ’, ‘ ∧ ’, and ‘ ¬ ’. How could you symbolise an exclusive or using only two connectives? Is there any way to symbolise an exclusive or using only one connective? For two connectives, we could offer any of the following: ¬ ( A ↔ B ) ( ¬ A ↔ B ) ( ¬ ( ¬ A ∧ ¬ B ) ∧ ¬ ( A ∧ B )) But if we wanted to symbolise it using only one connective, we would have to introduce a new primitive connective. Sentences of TFL 7 A. For each of the following: (a) Is it a sentence of TFL, strictly speaking? (b) Is it a sentence of TFL, allowing for our relaxed bracketing conventions? 1. ( A ) (a) no (b) no 2. J 374 ∨ ¬ J 374 (a) no (b) yes 3. ¬¬¬¬ F (a) yes (b) yes 4. ¬ ∧ S (a) no (b) no 5. ( G ∧ ¬ G ) (a) yes (b) yes 6. ( A → ( A ∧ ¬ F )) ∨ ( D ↔ E ) (a) no (b) yes 7. [( Z ↔ S ) → W ] ∧ [ J ∨ X ] (a) no (b) yes 8. ( F ↔ ¬ D → J ) ∨ ( C ∧ D ) (a) no (b) no B. Are there any sentences of TFL that contain no atomic sentences? Explain your answer. No. Atomic sentences contain atomic sentences (trivially). And every more complicated sentence is built up out of less complicated sentences, that were in turn built out of less complicated sentences, . . . , that were ultimately built out of atomic sentences. C. What is the scope of each connective in the sentence [ ( H → I ) ∨ ( I → H ) ] ∧ ( J ∨ K ) The scope of the left-most instance of ‘ → ’ is ‘( H → I )’. The scope of the right-most instance of ‘ → ’ is ‘( I → H )’. The scope of the left-most instance of ‘ ∨ is ‘ [ ( H → I ) ∨ ( I → H ) ] ’ The scope of the right-most instance of ‘ ∨ ’ is ‘( J ∨ K )’ The scope of the conjunction is the entire sentence; so conjunction is the main logical connective of the sentence. 8 Using truth tables 11 A. Complete truth tables for each of the following: 1. A → A A A → A T T T T F F T F 2. C → ¬ C C C → ¬ C T T F F T F F T T F 3. ( A ↔ B ) ↔ ¬ ( A ↔ ¬ B ) A B ( A ↔ B ) ↔ ¬ ( A ↔ ¬ B ) T T T T T T T T F F T T F T F F T F T T T F F T F F T T F F T F T F F F T F T T F F T F 4. ( A → B ) ∨ ( B → A ) A B ( A → B ) ∨ ( B → A ) T T T T T T T T T T F T F F T F T T F T F T T T T F F F F F T F T F T F 9 11. Using truth tables 10 5. ( A ∧ B ) → ( B ∨ A ) A B ( A ∧ B ) → ( B ∨ A ) T T T T T T T T T T F T F F T F T T F T F F T T T T F F F F F F T F F F 6. ¬ ( A ∨ B ) ↔ ( ¬ A ∧ ¬ B ) A B ¬ ( A ∨ B ) ↔ ( ¬ A ∧ ¬ B ) T T F T T T T F T F F T T F F T T F T F T F T F F T F F T T T T F F F T F F T F F F T T F TT F 7. [ ( A ∧ B ) ∧ ¬ ( A ∧ B ) ] ∧ C A B C [ ( A ∧ B ) ∧ ¬ ( A ∧ B ) ] ∧ C T T T T T T F F T T T F T T T F T T T F F T T T F F T F T T F F F T T F F F T T F F T F F F T T F F F F F T T F F T F T F F F F T F T F F F T F T F F T F F F F T F F F F T F F T F T F F F F F F F T F F F F F 8. [( A ∧ B ) ∧ C ] → B A B C [( A ∧ B ) ∧ C ] → B T T T T T T T T T T T T F T T T F F T T T F T T F F F T T F T F F T F F F F T F F T T F F T F T T T F T F F F T F F T T F F T F F F F T T F F F F F F F F F T F 11. Using truth tables 11 9. ¬ [ ( C ∨ A ) ∨ B ] A B C ¬ [ ( C ∨ A ) ∨ B ] T T T F T T T T T T T F F F T T T T T F T F T T T T F T F F F F T T T F F T T F T T F T T F T F F F F F T T F F T F T T F T F F F F T F F F F F B. Check all the claims made in introducing the new notational conventions in § 10.3, i.e. show that: 1. ‘(( A ∧ B ) ∧ C )’ and ‘( A ∧ ( B ∧ C ))’ have the same truth table A B C ( A ∧ B ) ∧ C A ∧ ( B ∧ C ) T T T T T T T T T T T T T T T F T T T F F T F T F F T F T T F F F T T F F F T T F F T F F F F T F F F F F T T F F T F T F F T T T F T F F F T F F F F T F F F F T F F F F T F F F F T F F F F F F F F F F F F F 2. ‘(( A ∨ B ) ∨ C )’ and ‘( A ∨ ( B ∨ C ))’ have the same truth table A B C ( A ∨ B ) ∨ C A ∨ ( B ∨ C ) T T T T T T T T T T T T T T T F T T T T F T T T T F T F T T T F T T T T F T T T F F T T F T F T T F F F F T T F T T T T F T T T T F T F F T T T F F T T T F F F T F F F T T F T F T T F F F F F F F F F F F F F 11. Using truth tables 12 3. ‘(( A ∨ B ) ∧ C )’ and ‘( A ∨ ( B ∧ C ))’ do not have the same truth table A B C ( A ∨ B ) ∧ C A ∨ ( B ∧ C ) T T T T T T T T T T T T T T T F T T T F F T T T F F T F T T T F T T T T F F T T F F T T F F F T T F F F F T T F T T T T F T T T T F T F F T T F F F F T F F F F T F F F F T F F F F T F F F F F F F F F F F F F 4. ‘(( A → B ) → C )’ and ‘( A → ( B → C ))’ do not have the same truth table A B C ( A → B ) → C A → ( B → C ) T T T T T T T T T T T T T T T F T T T F F T F T F F T F T T F F T T T T F T T T F F T F F T F T T F T F F T T F T T T T F T T T T F T F F T T F F F T T F F F F T F T F T T F T F T T F F F F T F F F F T F T F Also, check whether: 5. ‘(( A ↔ B ) ↔ C )’ and ‘( A ↔ ( B ↔ C ))’ have the same truth table Indeed they do: A B C ( A ↔ B ) ↔ C A ↔ ( B ↔ C ) T T T T T T T T T T T T T T T F T T T F F T F T F F T F T T F F F T T F F F T T F F T F F T F T T F T F F T T F F T F T F F T T T F T F F F T T F F T T F F F F T F T F T T F T F F T F F F F T F F F F F F T F Semantic concepts 12 A. Revisit your answers to § 10 A . Determine which sentences were tautologies, which were contradictions, and which were neither tautologies nor contradic- tions. 1. A → A Tautology 2. C → ¬ C Neither 3. ( A ↔ B ) ↔ ¬ ( A ↔ ¬ B ) Tautology 4. ( A → B ) ∨ ( B → A ) Tautology 5. ( A ∧ B ) → ( B ∨ A ) Tautology 6. ¬ ( A ∨ B ) ↔ ( ¬ A ∧ ¬ B ) Tautology 7. [ ( A ∧ B ) ∧ ¬ ( A ∧ B ) ] ∧ C Contradiction 8. [( A ∧ B ) ∧ C ] → B Tautology 9. ¬ [ ( C ∨ A ) ∨ B ] Neither B. Use truth tables to determine whether these sentences are jointly consistent, or jointly inconsistent: 1. A → A , ¬ A → ¬ A , A ∧ A , A ∨ A Jointly consistent (see line 1) A A → A ¬ A → ¬ A A ∧ A A ∨ A T T T T F T T F T T T T T T T F F T F T F T T F F F F F F F 2. A ∨ B , A → C , B → C Jointly consistent (see line 1) 13 12. Semantic concepts 14 A B C A ∨ B A → C B → C T T T T T T T T T T T T T T F T T T T F F T F F T F T T T T T T T F T T T F F T T F T F F F T F F T T F T F F T T T T T F T F F T T F T F T F F F F T F F F F T T F T T F F F F F F F T F F T F 3. B ∧ ( C ∨ A ), A → B , ¬ ( B ∨ C ) Jointly inconsistent A B C B ∧ ( C ∨ A ) A → B ¬ ( B ∨ C ) T T T T T T T T T T T F T T T T T F T T F T T T T T F T T F T F T F F T T T T F F F F T T T F F F F F T T T F F T F F F F T T T T T T F F T T F T T T F T F T F F F F F T T F T T F F F T F F T T F F T F F F T T F F F F F F F F F T F T F F F 4. A ↔ ( B ∨ C ), C → ¬ A , A → ¬ B Jointly consistent (see line 8) A B C A ↔ ( B ∨ C ) C → ¬ A A → ¬ B T T T T T T T T T F F T T F F T T T F T T T T F F T F T T F F T T F T T T F T T T F F T T T T F T F F T F F F F F T F T T T T F F T T F F T T T T T T F F T F T F T F F F T T F F T T F F T F T F F T F F F T T T T T F F T T F F F F F T F F F F T T F F T T F C. Use truth tables to determine whether each argument is valid or invalid. 1. A → A . ̇ . A Invalid (see line 2) 12. Semantic concepts 15 A A → A A T T T T T F F T F F 2. A → ( A ∧ ¬ A ) ̇ ¬ A Valid A A → ( A ∧ ¬ A ) ¬ A T T F T F F T F T F F T F F T F T F 3. A ∨ ( B → A ) ̇ ¬ A → ¬ B Valid A B A ∨ ( B → A ) ¬ A → ¬ B T T T T T T T F T T F T T F T T F T T F T T T F F T F F T F F T F F F T F F F T F T F T F T T F 4. A ∨ B, B ∨ C, ¬ A . ̇ . B ∧ C Invalid (see line 6) A B C A ∨ B B ∨ C ¬ A B ∧ C T T T T T T T T T F T T T T T T F T T T T T F F T T F F T F T T T F F T T F T F F T T F F T T F F F F F T F F F T T T F T T T T T T F T T T T T F F T T T T F T F T F F T F T F F F F T T T F F F T T F F F F F F F F T F F F F 5. ( B ∧ A ) → C, ( C ∧ A ) → B . ̇ ( C ∧ B ) → A Invalid (see line 5) A B C ( B ∧ A ) → C ( C ∧ A ) → B ( C ∧ B ) → A T T T T T T T T T T T T T T T T T T T T F T T T F F F F T T T F F T T T T F T F F T T T T T T F F T F F T T T F F F F T T F F F T T F F F F T T F T T T F F T T T F F T T T T T F F F T F T F F T F F F F T T F F T T F F F T F F F T T T F F T F T F F T F F F F F F F T F F F F T F F F F T F 12. Semantic concepts 16 D. Answer each of the questions below and justify your answer. 1. Suppose that A and B are tautologically equivalent. What can you say about A ↔ B ? A and B have the same truth value on every line of a truth table, so A ↔ B is true on every line. It is a tautology. 2. Suppose that ( A ∧ B ) → C is neither a tautology nor a contradiction. What can you say about whether A , B ̇ C is valid? Since the sentence ( A ∧ B ) → C is not a tautology, there is some line on which it is false. Since it is a conditional, on that line, A and B are true and C is false. So the argument is invalid. 3. Suppose that A , B and C are jointly tautologically inconsistent. What can you say about ( A ∧ B ∧ C )? Since the sentences are jointly tautologically inconsistent, there is no valuation on which they are all true. So their conjunction is false on every valuation. It is a contradiction 4. Suppose that A is a contradiction. What can you say about whether A , B C ? Since A is false on every line of a truth table, there is no line on which A and B are true and C is false. So the entailment holds. 5. Suppose that C is a tautology. What can you say about whether A , B C ? Since C is true on every line of a truth table, there is no line on which A and B are true and C is false. So the entailment holds. 6. Suppose that A and B are tautologically equivalent. What can you say about ( A ∨ B )? Not much. Since A and B are true on exactly the same lines of the truth table, their disjunction is true on exactly the same lines. So, their disjunction is tautologically equivalent to them. 7. Suppose that A and B are not tautologically equivalent. What can you say about ( A ∨ B )? A and B have different truth values on at least one line of a truth table, and ( A ∨ B ) will be true on that line. On other lines, it might be true or false. So ( A ∨ B ) is either a tautology or it is contingent; it is not a contradiction. 12. Semantic concepts 17 E. Consider the following principle: Suppose A and B are tautologically equivalent. Suppose an argument contains A (either as a premise, or as the conclusion). The validity of the argument would be unaffected, if we replaced A with B Is this principle correct? Explain your answer. The principle is correct. Since A and B are tautologically equivalent, they have the same truth table. So every valuation that makes A true also makes B true, and every valuation that makes A false also makes B false. So if no valuation makes all the premises true and the conclusion false, when A was among the premises or the conclusion, then no valuation makes all the premises true and the conclusion false, when we replace A with B